If for a distribution $\Sigma(x-5)=3, \Sigma(x-5)^{2}=43$ and the total number of item is $18,$ find the mean and standard deviation.
If for a distribution $\Sigma(x-5)=3, \Sigma(x-5)^{2}=43$ and the total number of item is $18,$ find the mean and standard deviation.
Given, $n=18, \Sigma(x-5)=3$ and $\Sigma(x-5)^{2}=43$
$\therefore \quad \operatorname{Mean}=A+\frac{\Sigma(x-5)}{18}=5+\frac{3}{18}=5+0.1666=5.1666=5.17$
and $\quad SD =\sqrt{\frac{\Sigma(x-5)^{2}}{n}-\left(\frac{\Sigma(x-5)}{n}\right)^{2}}=\sqrt{\frac{43}{18}-\left(\frac{3}{18}\right)^{2}}$
$=\sqrt{2.3889-(0.166)^{2}}=\sqrt{2.3889-0.0277}=1.53$
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